Optimal. Leaf size=266 \[ -\frac{16}{49 \sqrt{2 x+1}}-\frac{4}{21 (2 x+1)^{3/2}}-\frac{1}{49} \sqrt{\frac{1}{434} \left (1225 \sqrt{35}-7162\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{49} \sqrt{\frac{1}{434} \left (1225 \sqrt{35}-7162\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{49} \sqrt{\frac{2}{217} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-\frac{1}{49} \sqrt{\frac{2}{217} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]
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Rubi [A] time = 0.382157, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {709, 828, 826, 1169, 634, 618, 204, 628} \[ -\frac{16}{49 \sqrt{2 x+1}}-\frac{4}{21 (2 x+1)^{3/2}}-\frac{1}{49} \sqrt{\frac{1}{434} \left (1225 \sqrt{35}-7162\right )} \log \left (5 (2 x+1)-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{49} \sqrt{\frac{1}{434} \left (1225 \sqrt{35}-7162\right )} \log \left (5 (2 x+1)+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{2 x+1}+\sqrt{35}\right )+\frac{1}{49} \sqrt{\frac{2}{217} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{\sqrt{10 \left (2+\sqrt{35}\right )}-10 \sqrt{2 x+1}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right )-\frac{1}{49} \sqrt{\frac{2}{217} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\frac{10 \sqrt{2 x+1}+\sqrt{10 \left (2+\sqrt{35}\right )}}{\sqrt{10 \left (\sqrt{35}-2\right )}}\right ) \]
Antiderivative was successfully verified.
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Rule 709
Rule 828
Rule 826
Rule 1169
Rule 634
Rule 618
Rule 204
Rule 628
Rubi steps
\begin{align*} \int \frac{1}{(1+2 x)^{5/2} \left (2+3 x+5 x^2\right )} \, dx &=-\frac{4}{21 (1+2 x)^{3/2}}+\frac{1}{7} \int \frac{-1-10 x}{(1+2 x)^{3/2} \left (2+3 x+5 x^2\right )} \, dx\\ &=-\frac{4}{21 (1+2 x)^{3/2}}-\frac{16}{49 \sqrt{1+2 x}}+\frac{1}{49} \int \frac{-39-40 x}{\sqrt{1+2 x} \left (2+3 x+5 x^2\right )} \, dx\\ &=-\frac{4}{21 (1+2 x)^{3/2}}-\frac{16}{49 \sqrt{1+2 x}}+\frac{2}{49} \operatorname{Subst}\left (\int \frac{-38-40 x^2}{7-4 x^2+5 x^4} \, dx,x,\sqrt{1+2 x}\right )\\ &=-\frac{4}{21 (1+2 x)^{3/2}}-\frac{16}{49 \sqrt{1+2 x}}+\frac{\operatorname{Subst}\left (\int \frac{-38 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-\left (-38+8 \sqrt{35}\right ) x}{\sqrt{\frac{7}{5}}-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )}{49 \sqrt{14 \left (2+\sqrt{35}\right )}}+\frac{\operatorname{Subst}\left (\int \frac{-38 \sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+\left (-38+8 \sqrt{35}\right ) x}{\sqrt{\frac{7}{5}}+\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )}{49 \sqrt{14 \left (2+\sqrt{35}\right )}}\\ &=-\frac{4}{21 (1+2 x)^{3/2}}-\frac{16}{49 \sqrt{1+2 x}}-\frac{\left (140+19 \sqrt{35}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{\frac{7}{5}}-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )}{1715}-\frac{\left (140+19 \sqrt{35}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{\frac{7}{5}}+\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )}{1715}-\frac{1}{49} \sqrt{\frac{1}{434} \left (-7162+1225 \sqrt{35}\right )} \operatorname{Subst}\left (\int \frac{-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 x}{\sqrt{\frac{7}{5}}-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )+\frac{1}{49} \sqrt{\frac{1}{434} \left (-7162+1225 \sqrt{35}\right )} \operatorname{Subst}\left (\int \frac{\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 x}{\sqrt{\frac{7}{5}}+\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )} x+x^2} \, dx,x,\sqrt{1+2 x}\right )\\ &=-\frac{4}{21 (1+2 x)^{3/2}}-\frac{16}{49 \sqrt{1+2 x}}-\frac{1}{49} \sqrt{\frac{1}{434} \left (-7162+1225 \sqrt{35}\right )} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )+\frac{1}{49} \sqrt{\frac{1}{434} \left (-7162+1225 \sqrt{35}\right )} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )+\frac{\left (2 \left (140+19 \sqrt{35}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{2}{5} \left (2-\sqrt{35}\right )-x^2} \, dx,x,-\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 \sqrt{1+2 x}\right )}{1715}+\frac{\left (2 \left (140+19 \sqrt{35}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{2}{5} \left (2-\sqrt{35}\right )-x^2} \, dx,x,\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 \sqrt{1+2 x}\right )}{1715}\\ &=-\frac{4}{21 (1+2 x)^{3/2}}-\frac{16}{49 \sqrt{1+2 x}}+\frac{1}{49} \sqrt{\frac{2}{217} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\sqrt{\frac{5}{2 \left (-2+\sqrt{35}\right )}} \left (\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}-2 \sqrt{1+2 x}\right )\right )-\frac{1}{49} \sqrt{\frac{2}{217} \left (7162+1225 \sqrt{35}\right )} \tan ^{-1}\left (\sqrt{\frac{5}{2 \left (-2+\sqrt{35}\right )}} \left (\sqrt{\frac{2}{5} \left (2+\sqrt{35}\right )}+2 \sqrt{1+2 x}\right )\right )-\frac{1}{49} \sqrt{\frac{1}{434} \left (-7162+1225 \sqrt{35}\right )} \log \left (\sqrt{35}-\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )+\frac{1}{49} \sqrt{\frac{1}{434} \left (-7162+1225 \sqrt{35}\right )} \log \left (\sqrt{35}+\sqrt{10 \left (2+\sqrt{35}\right )} \sqrt{1+2 x}+5 (1+2 x)\right )\\ \end{align*}
Mathematica [C] time = 0.527477, size = 133, normalized size = 0.5 \[ \frac{2 \left (-\frac{2170 (24 x+19)}{(2 x+1)^{3/2}}+3 i \sqrt{10-5 i \sqrt{31}} \left (178 \sqrt{31}+589 i\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2-i \sqrt{31}}}\right )-3 i \sqrt{10+5 i \sqrt{31}} \left (178 \sqrt{31}-589 i\right ) \tanh ^{-1}\left (\frac{\sqrt{10 x+5}}{\sqrt{2+i \sqrt{31}}}\right )\right )}{159495} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.075, size = 625, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (5 \, x^{2} + 3 \, x + 2\right )}{\left (2 \, x + 1\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.7599, size = 2337, normalized size = 8.79 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (2 x + 1\right )^{\frac{5}{2}} \left (5 x^{2} + 3 x + 2\right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (5 \, x^{2} + 3 \, x + 2\right )}{\left (2 \, x + 1\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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